Subject: Re: [xsl] How do you select all unique first-position characters? From: David Carlisle <davidc@xxxxxxxxx> Date: Thu, 21 Nov 2002 15:38:36 GMT |
Could you explain to me why the union of the current node with that of the first node returned from the xsl:key is needed in the count? Wouldn't that always be 1? This is the "standard trick" for determining if two nodes are the same node (as opposed to having the same value) If you want to know if . and key(,,,)[1] are the same thing then you can form . | key(,,,)[1] this is a node set consisting of 1 node if they are the same and two if they are different (actually it would also have size 1 if key() did not return anything, but from the context you know that there will be something with that key in this case) you can do generate-id(.) = generate-id(key(,,,)[1]) which does the same test. In Xpath 2 (if you can cope with the schema-horrors) then you can just test for node equality directly with . is key(,,,)[1] David _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Scanning Service. For further information visit http://www.star.net.uk/stats.asp or alternatively call Star Internet for details on the Virus Scanning Service. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] How do you select all uni, Jeff Beadle | Thread | RE: [xsl] How do you select all uni, Wendell Piez |
Re: [xsl] Exclusions in XPATH, David Carlisle | Date | RE: [xsl] How do you select all uni, Wendell Piez |
Month |